3.7.21 \(\int \frac {(a+b x)^{3/2}}{x^3 (c+d x)^{5/2}} \, dx\)

Optimal. Leaf size=204 \[ -\frac {\left (35 a^2 d^2-30 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a} c^{9/2}}-\frac {5 d \sqrt {a+b x} (11 b c-21 a d)}{12 c^4 \sqrt {c+d x}}-\frac {d \sqrt {a+b x} (23 b c-35 a d)}{12 c^3 (c+d x)^{3/2}}-\frac {\sqrt {a+b x} (5 b c-7 a d)}{4 c^2 x (c+d x)^{3/2}}-\frac {a \sqrt {a+b x}}{2 c x^2 (c+d x)^{3/2}} \]

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Rubi [A]  time = 0.23, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {98, 151, 152, 12, 93, 208} \begin {gather*} -\frac {\left (35 a^2 d^2-30 a b c d+3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a} c^{9/2}}-\frac {5 d \sqrt {a+b x} (11 b c-21 a d)}{12 c^4 \sqrt {c+d x}}-\frac {d \sqrt {a+b x} (23 b c-35 a d)}{12 c^3 (c+d x)^{3/2}}-\frac {\sqrt {a+b x} (5 b c-7 a d)}{4 c^2 x (c+d x)^{3/2}}-\frac {a \sqrt {a+b x}}{2 c x^2 (c+d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(3/2)/(x^3*(c + d*x)^(5/2)),x]

[Out]

-(d*(23*b*c - 35*a*d)*Sqrt[a + b*x])/(12*c^3*(c + d*x)^(3/2)) - (a*Sqrt[a + b*x])/(2*c*x^2*(c + d*x)^(3/2)) -
((5*b*c - 7*a*d)*Sqrt[a + b*x])/(4*c^2*x*(c + d*x)^(3/2)) - (5*d*(11*b*c - 21*a*d)*Sqrt[a + b*x])/(12*c^4*Sqrt
[c + d*x]) - ((3*b^2*c^2 - 30*a*b*c*d + 35*a^2*d^2)*ArcTanh[(Sqrt[c]*Sqrt[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])])/
(4*Sqrt[a]*c^(9/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 98

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c -
a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^(p + 1))/(b*(b*e - a*f)*(m + 1)), x] + Dist[1/(b*(b*e - a*
f)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d
*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /;
FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p
] || IntegersQ[p, m + n])

Rule 151

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegerQ[m]

Rule 152

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)), x_Symb
ol] :> Simp[((b*g - a*h)*(a + b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*
f)), x] + Dist[1/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*x)^p*Simp[(a*d*f*
g - b*(d*e + c*f)*g + b*c*e*h)*(m + 1) - (b*g - a*h)*(d*e*(n + 1) + c*f*(p + 1)) - d*f*(b*g - a*h)*(m + n + p
+ 3)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*n, 2*p]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin {align*} \int \frac {(a+b x)^{3/2}}{x^3 (c+d x)^{5/2}} \, dx &=-\frac {a \sqrt {a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac {\int \frac {-\frac {1}{2} a (5 b c-7 a d)-b (2 b c-3 a d) x}{x^2 \sqrt {a+b x} (c+d x)^{5/2}} \, dx}{2 c}\\ &=-\frac {a \sqrt {a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac {(5 b c-7 a d) \sqrt {a+b x}}{4 c^2 x (c+d x)^{3/2}}+\frac {\int \frac {\frac {1}{4} a \left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right )-a b d (5 b c-7 a d) x}{x \sqrt {a+b x} (c+d x)^{5/2}} \, dx}{2 a c^2}\\ &=-\frac {d (23 b c-35 a d) \sqrt {a+b x}}{12 c^3 (c+d x)^{3/2}}-\frac {a \sqrt {a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac {(5 b c-7 a d) \sqrt {a+b x}}{4 c^2 x (c+d x)^{3/2}}-\frac {\int \frac {-\frac {3}{8} a (b c-a d) \left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right )+\frac {1}{4} a b d (23 b c-35 a d) (b c-a d) x}{x \sqrt {a+b x} (c+d x)^{3/2}} \, dx}{3 a c^3 (b c-a d)}\\ &=-\frac {d (23 b c-35 a d) \sqrt {a+b x}}{12 c^3 (c+d x)^{3/2}}-\frac {a \sqrt {a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac {(5 b c-7 a d) \sqrt {a+b x}}{4 c^2 x (c+d x)^{3/2}}-\frac {5 d (11 b c-21 a d) \sqrt {a+b x}}{12 c^4 \sqrt {c+d x}}+\frac {2 \int \frac {3 a (b c-a d)^2 \left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right )}{16 x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{3 a c^4 (b c-a d)^2}\\ &=-\frac {d (23 b c-35 a d) \sqrt {a+b x}}{12 c^3 (c+d x)^{3/2}}-\frac {a \sqrt {a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac {(5 b c-7 a d) \sqrt {a+b x}}{4 c^2 x (c+d x)^{3/2}}-\frac {5 d (11 b c-21 a d) \sqrt {a+b x}}{12 c^4 \sqrt {c+d x}}+\frac {\left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right ) \int \frac {1}{x \sqrt {a+b x} \sqrt {c+d x}} \, dx}{8 c^4}\\ &=-\frac {d (23 b c-35 a d) \sqrt {a+b x}}{12 c^3 (c+d x)^{3/2}}-\frac {a \sqrt {a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac {(5 b c-7 a d) \sqrt {a+b x}}{4 c^2 x (c+d x)^{3/2}}-\frac {5 d (11 b c-21 a d) \sqrt {a+b x}}{12 c^4 \sqrt {c+d x}}+\frac {\left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right ) \operatorname {Subst}\left (\int \frac {1}{-a+c x^2} \, dx,x,\frac {\sqrt {a+b x}}{\sqrt {c+d x}}\right )}{4 c^4}\\ &=-\frac {d (23 b c-35 a d) \sqrt {a+b x}}{12 c^3 (c+d x)^{3/2}}-\frac {a \sqrt {a+b x}}{2 c x^2 (c+d x)^{3/2}}-\frac {(5 b c-7 a d) \sqrt {a+b x}}{4 c^2 x (c+d x)^{3/2}}-\frac {5 d (11 b c-21 a d) \sqrt {a+b x}}{12 c^4 \sqrt {c+d x}}-\frac {\left (3 b^2 c^2-30 a b c d+35 a^2 d^2\right ) \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )}{4 \sqrt {a} c^{9/2}}\\ \end {align*}

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Mathematica [A]  time = 0.22, size = 171, normalized size = 0.84 \begin {gather*} -\frac {-x^2 \left (35 a^2 d^2-30 a b c d+3 b^2 c^2\right ) \left (\sqrt {c} \sqrt {a+b x} (4 a c+3 a d x+b c x)-3 a^{3/2} (c+d x)^{3/2} \tanh ^{-1}\left (\frac {\sqrt {c} \sqrt {a+b x}}{\sqrt {a} \sqrt {c+d x}}\right )\right )+3 c^{5/2} x (a+b x)^{5/2} (b c-7 a d)+6 a c^{7/2} (a+b x)^{5/2}}{12 a^2 c^{9/2} x^2 (c+d x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(3/2)/(x^3*(c + d*x)^(5/2)),x]

[Out]

-1/12*(6*a*c^(7/2)*(a + b*x)^(5/2) + 3*c^(5/2)*(b*c - 7*a*d)*x*(a + b*x)^(5/2) - (3*b^2*c^2 - 30*a*b*c*d + 35*
a^2*d^2)*x^2*(Sqrt[c]*Sqrt[a + b*x]*(4*a*c + b*c*x + 3*a*d*x) - 3*a^(3/2)*(c + d*x)^(3/2)*ArcTanh[(Sqrt[c]*Sqr
t[a + b*x])/(Sqrt[a]*Sqrt[c + d*x])]))/(a^2*c^(9/2)*x^2*(c + d*x)^(3/2))

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IntegrateAlgebraic [A]  time = 0.31, size = 295, normalized size = 1.45 \begin {gather*} \frac {\left (-35 a^2 d^2+30 a b c d-3 b^2 c^2\right ) \tanh ^{-1}\left (\frac {\sqrt {a} \sqrt {c+d x}}{\sqrt {c} \sqrt {a+b x}}\right )}{4 \sqrt {a} c^{9/2}}+\frac {(a+b x)^{3/2} \left (\frac {105 a^3 d^2 (c+d x)^3}{(a+b x)^3}-\frac {175 a^2 c d^2 (c+d x)^2}{(a+b x)^2}-\frac {90 a^2 b c d (c+d x)^3}{(a+b x)^3}-\frac {15 b^2 c^3 (c+d x)^2}{(a+b x)^2}+\frac {9 a b^2 c^2 (c+d x)^3}{(a+b x)^3}-\frac {48 b c^3 d (c+d x)}{a+b x}+\frac {56 a c^2 d^2 (c+d x)}{a+b x}+\frac {150 a b c^2 d (c+d x)^2}{(a+b x)^2}+8 c^3 d^2\right )}{12 c^4 (c+d x)^{3/2} \left (c-\frac {a (c+d x)}{a+b x}\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(3/2)/(x^3*(c + d*x)^(5/2)),x]

[Out]

((a + b*x)^(3/2)*(8*c^3*d^2 - (48*b*c^3*d*(c + d*x))/(a + b*x) + (56*a*c^2*d^2*(c + d*x))/(a + b*x) - (15*b^2*
c^3*(c + d*x)^2)/(a + b*x)^2 + (150*a*b*c^2*d*(c + d*x)^2)/(a + b*x)^2 - (175*a^2*c*d^2*(c + d*x)^2)/(a + b*x)
^2 + (9*a*b^2*c^2*(c + d*x)^3)/(a + b*x)^3 - (90*a^2*b*c*d*(c + d*x)^3)/(a + b*x)^3 + (105*a^3*d^2*(c + d*x)^3
)/(a + b*x)^3))/(12*c^4*(c + d*x)^(3/2)*(c - (a*(c + d*x))/(a + b*x))^2) + ((-3*b^2*c^2 + 30*a*b*c*d - 35*a^2*
d^2)*ArcTanh[(Sqrt[a]*Sqrt[c + d*x])/(Sqrt[c]*Sqrt[a + b*x])])/(4*Sqrt[a]*c^(9/2))

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fricas [A]  time = 10.42, size = 634, normalized size = 3.11 \begin {gather*} \left [\frac {3 \, {\left ({\left (3 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 35 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (3 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{3} + {\left (3 \, b^{2} c^{4} - 30 \, a b c^{3} d + 35 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {a c} \log \left (\frac {8 \, a^{2} c^{2} + {\left (b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2}\right )} x^{2} - 4 \, {\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {a c} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (a b c^{2} + a^{2} c d\right )} x}{x^{2}}\right ) - 4 \, {\left (6 \, a^{2} c^{4} + 5 \, {\left (11 \, a b c^{2} d^{2} - 21 \, a^{2} c d^{3}\right )} x^{3} + 2 \, {\left (39 \, a b c^{3} d - 70 \, a^{2} c^{2} d^{2}\right )} x^{2} + 3 \, {\left (5 \, a b c^{4} - 7 \, a^{2} c^{3} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, {\left (a c^{5} d^{2} x^{4} + 2 \, a c^{6} d x^{3} + a c^{7} x^{2}\right )}}, \frac {3 \, {\left ({\left (3 \, b^{2} c^{2} d^{2} - 30 \, a b c d^{3} + 35 \, a^{2} d^{4}\right )} x^{4} + 2 \, {\left (3 \, b^{2} c^{3} d - 30 \, a b c^{2} d^{2} + 35 \, a^{2} c d^{3}\right )} x^{3} + {\left (3 \, b^{2} c^{4} - 30 \, a b c^{3} d + 35 \, a^{2} c^{2} d^{2}\right )} x^{2}\right )} \sqrt {-a c} \arctan \left (\frac {{\left (2 \, a c + {\left (b c + a d\right )} x\right )} \sqrt {-a c} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (a b c d x^{2} + a^{2} c^{2} + {\left (a b c^{2} + a^{2} c d\right )} x\right )}}\right ) - 2 \, {\left (6 \, a^{2} c^{4} + 5 \, {\left (11 \, a b c^{2} d^{2} - 21 \, a^{2} c d^{3}\right )} x^{3} + 2 \, {\left (39 \, a b c^{3} d - 70 \, a^{2} c^{2} d^{2}\right )} x^{2} + 3 \, {\left (5 \, a b c^{4} - 7 \, a^{2} c^{3} d\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{24 \, {\left (a c^{5} d^{2} x^{4} + 2 \, a c^{6} d x^{3} + a c^{7} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^3/(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

[1/48*(3*((3*b^2*c^2*d^2 - 30*a*b*c*d^3 + 35*a^2*d^4)*x^4 + 2*(3*b^2*c^3*d - 30*a*b*c^2*d^2 + 35*a^2*c*d^3)*x^
3 + (3*b^2*c^4 - 30*a*b*c^3*d + 35*a^2*c^2*d^2)*x^2)*sqrt(a*c)*log((8*a^2*c^2 + (b^2*c^2 + 6*a*b*c*d + a^2*d^2
)*x^2 - 4*(2*a*c + (b*c + a*d)*x)*sqrt(a*c)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(a*b*c^2 + a^2*c*d)*x)/x^2) - 4*(6
*a^2*c^4 + 5*(11*a*b*c^2*d^2 - 21*a^2*c*d^3)*x^3 + 2*(39*a*b*c^3*d - 70*a^2*c^2*d^2)*x^2 + 3*(5*a*b*c^4 - 7*a^
2*c^3*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^5*d^2*x^4 + 2*a*c^6*d*x^3 + a*c^7*x^2), 1/24*(3*((3*b^2*c^2*d^2
- 30*a*b*c*d^3 + 35*a^2*d^4)*x^4 + 2*(3*b^2*c^3*d - 30*a*b*c^2*d^2 + 35*a^2*c*d^3)*x^3 + (3*b^2*c^4 - 30*a*b*c
^3*d + 35*a^2*c^2*d^2)*x^2)*sqrt(-a*c)*arctan(1/2*(2*a*c + (b*c + a*d)*x)*sqrt(-a*c)*sqrt(b*x + a)*sqrt(d*x +
c)/(a*b*c*d*x^2 + a^2*c^2 + (a*b*c^2 + a^2*c*d)*x)) - 2*(6*a^2*c^4 + 5*(11*a*b*c^2*d^2 - 21*a^2*c*d^3)*x^3 + 2
*(39*a*b*c^3*d - 70*a^2*c^2*d^2)*x^2 + 3*(5*a*b*c^4 - 7*a^2*c^3*d)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a*c^5*d^2*
x^4 + 2*a*c^6*d*x^3 + a*c^7*x^2)]

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giac [B]  time = 11.76, size = 1234, normalized size = 6.05

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^3/(d*x+c)^(5/2),x, algorithm="giac")

[Out]

-2/3*sqrt(b*x + a)*((5*b^5*c^6*d^3*abs(b) - 14*a*b^4*c^5*d^4*abs(b) + 9*a^2*b^3*c^4*d^5*abs(b))*(b*x + a)/(b^3
*c^9*d - a*b^2*c^8*d^2) + 3*(2*b^6*c^7*d^2*abs(b) - 7*a*b^5*c^6*d^3*abs(b) + 8*a^2*b^4*c^5*d^4*abs(b) - 3*a^3*
b^3*c^4*d^5*abs(b))/(b^3*c^9*d - a*b^2*c^8*d^2))/(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) - 1/4*(3*sqrt(b*d)*b^4*
c^2 - 30*sqrt(b*d)*a*b^3*c*d + 35*sqrt(b*d)*a^2*b^2*d^2)*arctan(-1/2*(b^2*c + a*b*d - (sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(sqrt(-a*b*c*d)*b))/(sqrt(-a*b*c*d)*b*c^4*abs(b)) - 1/2*(5*sqrt(b*d
)*b^10*c^5 - 31*sqrt(b*d)*a*b^9*c^4*d + 74*sqrt(b*d)*a^2*b^8*c^3*d^2 - 86*sqrt(b*d)*a^3*b^7*c^2*d^3 + 49*sqrt(
b*d)*a^4*b^6*c*d^4 - 11*sqrt(b*d)*a^5*b^5*d^5 - 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)
*b*d - a*b*d))^2*b^8*c^4 + 52*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a*b^
7*c^3*d - 26*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^2*b^6*c^2*d^2 - 44*
sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^5*c*d^3 + 33*sqrt(b*d)*(sqrt
(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^4*d^4 + 15*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x +
a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*b^6*c^3 - 23*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*
x + a)*b*d - a*b*d))^4*a*b^5*c^2*d - 7*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d
))^4*a^2*b^4*c*d^2 - 33*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^3*b^3*d^
3 - 5*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*b^4*c^2 + 2*sqrt(b*d)*(sqrt(
b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a*b^3*c*d + 11*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a)
- sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^6*a^2*b^2*d^2)/((b^4*c^2 - 2*a*b^3*c*d + a^2*b^2*d^2 - 2*(sqrt(b*d)*sqr
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*b^2*c - 2*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a
)*b*d - a*b*d))^2*a*b*d + (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^4)^2*c^4*abs(b))

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maple [B]  time = 0.04, size = 679, normalized size = 3.33 \begin {gather*} -\frac {\sqrt {b x +a}\, \left (105 a^{2} d^{4} x^{4} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-90 a b c \,d^{3} x^{4} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+9 b^{2} c^{2} d^{2} x^{4} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+210 a^{2} c \,d^{3} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-180 a b \,c^{2} d^{2} x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+18 b^{2} c^{3} d \,x^{3} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+105 a^{2} c^{2} d^{2} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-90 a b \,c^{3} d \,x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )+9 b^{2} c^{4} x^{2} \ln \left (\frac {a d x +b c x +2 a c +2 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}}{x}\right )-210 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,d^{3} x^{3}+110 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b c \,d^{2} x^{3}-280 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a c \,d^{2} x^{2}+156 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b \,c^{2} d \,x^{2}-42 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,c^{2} d x +30 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, b \,c^{3} x +12 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, a \,c^{3}\right )}{24 \sqrt {a c}\, \sqrt {\left (b x +a \right ) \left (d x +c \right )}\, \left (d x +c \right )^{\frac {3}{2}} c^{4} x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(3/2)/x^3/(d*x+c)^(5/2),x)

[Out]

-1/24*(b*x+a)^(1/2)/c^4*(105*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^4*a^2*d^4-90*ln
((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^4*a*b*c*d^3+9*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1
/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^4*b^2*c^2*d^2+210*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2
))/x)*x^3*a^2*c*d^3-180*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*a*b*c^2*d^2+18*ln(
(a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^3*b^2*c^3*d+105*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(
1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*a^2*c^2*d^2-90*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2
))/x)*x^2*a*b*c^3*d+9*ln((a*d*x+b*c*x+2*a*c+2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2))/x)*x^2*b^2*c^4-210*x^3*a*d^
3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+110*x^3*b*c*d^2*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-280*x^2*a*c*d^2*(a*c
)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+156*x^2*b*c^2*d*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-42*x*a*c^2*d*(a*c)^(1/2)*(
(b*x+a)*(d*x+c))^(1/2)+30*x*b*c^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+12*a*c^3*(a*c)^(1/2)*((b*x+a)*(d*x+c))^(
1/2))/(a*c)^(1/2)/x^2/((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(3/2)

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(3/2)/x^3/(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for
more details)Is a*d-b*c zero or nonzero?

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (a+b\,x\right )}^{3/2}}{x^3\,{\left (c+d\,x\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(3/2)/(x^3*(c + d*x)^(5/2)),x)

[Out]

int((a + b*x)^(3/2)/(x^3*(c + d*x)^(5/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(3/2)/x**3/(d*x+c)**(5/2),x)

[Out]

Timed out

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